Is |x| < 1 ?

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lalmanistl: Junior | Next Rank: 30 Posts; Posts: 13; Joined: Wed Oct 01, 2008 4:41 pm; Location: USA

by lalmanistl » Mon Jan 25, 2010 9:10 pm

Is |x| < 1 ?

(1) |x + 1| = 2|x minus 1|

(2) |x minus 3| > 0

please help me to solve this question.

thanks in advance.

Last edited by lalmanistl on Mon Jan 25, 2010 9:39 pm, edited 2 times in total.

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Source: Beat The GMAT — Data Sufficiency | Mon Jan 25, 2010 9:10 pm

fibbonnaci: MBA Student; Posts: 403; Joined: Tue Dec 22, 2009 7:32 pmFollowed by: 22 members; Thanked: 98 times

by fibbonnaci » Mon Jan 25, 2010 9:34 pm

please post the question properly. the mathematical symbols are not clear.

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papgust: Community Manager; Posts: 1537; Joined: Mon Aug 10, 2009 6:10 pmFollowed by: 252 members; Thanked: 653 times

by papgust » Mon Jan 25, 2010 10:54 pm

lalmanistl wrote:Is |x| < 1 ?

(1) |x + 1| = 2|x minus 1|

(2) |x minus 3| > 0

Is -1 < x < 1?

(1) |x + 1| = 2|x - 1|

Square both sides,
(x+1)^2 = 4 (x-1)^2
x^2 + 2x + 1 = 4x^2 - 8x + 4
3x^2 - 10x + 3 = 0
x=3,1/3
1/3 lies b/w -1 and 1 But 3 does not lie b/w -1 and 1.
Insufficient.

(2) |x - 3| > 0
x - 3 > 0 ; -x + 3 > 0
x > 3; x < 3
x = 3.
x does not lie b/w -1 and 1. Sufficient.

Hence B

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VikingWarrior: Senior | Next Rank: 100 Posts; Posts: 89; Joined: Sat Jan 23, 2010 9:55 am; Thanked: 6 times

by VikingWarrior » Tue Jan 26, 2010 12:02 am

Papgust, I differ in my solution.
My answer would be C.
As you have correctky pointed out that from 1. we can derive x=3 or x=1/3
from 2 we can derive that x is not=3 i.e. for absolute value of (x-3) to be more than 0 x can be any no. but 3.
Thus combining 1 and 2 we get -1<x<1

Of course the assumption is that x is an integer

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tnguyen: Newbie | Next Rank: 10 Posts; Posts: 8; Joined: Sat Jan 23, 2010 1:59 am

by tnguyen » Tue Jan 26, 2010 1:00 am

can someone explain why squaring both sides is the first and correct step?

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ajith: Legendary Member; Posts: 1275; Joined: Thu Sep 21, 2006 11:13 pmFollowed by: 2 members; Location: Arabian Sea; Thanked: 125 times

by ajith » Tue Jan 26, 2010 1:28 am

papgust wrote:
(2) |x - 3| > 0
x - 3 > 0 ; -x + 3 > 0
x > 3; x < 3
x = 3.
x does not lie b/w -1 and 1. Sufficient.

Hence B

|x-3| is always positive no matter what the value of x is so, I do not think the conclusion that x do not lie between 1 and -1 is true

Hence E - Both statement together is not sufficient

Always borrow money from a pessimist, he doesn't expect to be paid back.

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ajith: Legendary Member; Posts: 1275; Joined: Thu Sep 21, 2006 11:13 pmFollowed by: 2 members; Location: Arabian Sea; Thanked: 125 times

by ajith » Tue Jan 26, 2010 1:33 am

tnguyen wrote:can someone explain why squaring both sides is the first and correct step?

Because of the mode operation.

say if x is -ve and y is +ve

|x| = |y| means -x = y

if x is -ve and y is -ve

it means

-x = -y

So we cannot be sure whether x = y or x = -y (if we have no idea about the signs of x and y)

but squaring solves that problem and makes both side positive

x^2 = y^2 (no matter what the signs of x and y are)

Always borrow money from a pessimist, he doesn't expect to be paid back.

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VikingWarrior: Senior | Next Rank: 100 Posts; Posts: 89; Joined: Sat Jan 23, 2010 9:55 am; Thanked: 6 times

by VikingWarrior » Tue Jan 26, 2010 2:02 am

Ajith,

|x-y|>0 for all integers x and y provided x is not equal to y
Absolute value cannot have a negative sign.

|x-3|>0 is only suggesting that x-3 cannot be equal to 0 therefore x is not equal to 3. Add this information to statement 1 and you have the answer.

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ajith: Legendary Member; Posts: 1275; Joined: Thu Sep 21, 2006 11:13 pmFollowed by: 2 members; Location: Arabian Sea; Thanked: 125 times

by ajith » Tue Jan 26, 2010 2:08 am

VikingWarrior wrote: |x-3|>0 is only suggesting that x-3 cannot be equal to 0 therefore x is not equal to 3. Add this information to statement 1 and you have the answer.

Agree there!

Always borrow money from a pessimist, he doesn't expect to be paid back.

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